pretty easy for a #6?

I would just like to mention that the lemma used by tastymath75025 can be generalized, and perhaps used to solve this problem for more general $n.$ Generalization: Let $(G, \cdot)$ be a group of order $n$ and let $k$ be an integer relatively prime to $n.$ Then the map $x \mapsto x^k$ is surjective. Proof: Choose any $g \in G.$ We must find an $x$ such that $x^k = g.$ Since $(n, k) = 1$, by Bézout's Lemma, there exist integers $\alpha, \beta$ satisfying $\alpha n + \beta k = 1.$ Set $x = g^{\beta}.$ Then by Lagrange's Theorem, \[ x^k = g^{\beta k} = \left(g^n\right)^\alpha \cdot g^{\beta k} = g^{\alpha n + \beta k} = g. \; \blacksquare \] Now, in the given problem, choose $p$ to be the largest prime divisor of $n.$ By the Chinese Remainder Theorem, it's enough to solve the congruence \[ p^2 + p \cdot m^p \equiv 0 \pmod{q^{\alpha}} \qquad (\star)\]for an arbitrary prime power $q^{\alpha}$ dividing $n.$ Unfortunately, $(\star)$ might not have a solution when $q = p.$ However, if $q \ne p$, the above result guarantees the existence of a solution. Let $G = (\mathbb{Z} / q^{\alpha} \mathbb{Z})^{\times}$ be the multiplicative group of integers modulo $q^{\alpha}.$ Then $\left(-p, q^{\alpha}\right) = 1$, so $-p \in G.$ Moreover, $G$ has order $\phi\left(q^{\alpha}\right) = q^{\alpha - 1}(q - 1)$, which is relatively prime to $p$ (since $p > q$). Therefore, the above result implies the existence of an integer $m$ satisfying $m^p \equiv -p \pmod{q^{\alpha}}.$ Clearly such an $m$ is also a solution to $(\star).$

http://artofproblemsolving.com/community/c6h1207294p5967343

Mine is nearly the same as #2. Let $n = p\cdot p_1 \cdot p_2 \cdot \dots p_k$ with $p > p_1 > p_2 > \dots > p_k$. We will prove that there exist $m \in \mathbb{Z}$ such that $p + m^p \equiv 0 \pmod{p_i}$ for all $1 \leq i \leq k$. We will prove that for fixed $1 \leq i \leq k$, there exist $m \in \mathbb{Z}$ such that the statement holds, and we will be done by Chinese remainder theorem. Let $g$ be a primitive root modulo $p_i$. Since $\gcd{(p, p_i - 1)} = 1$, as $\alpha$ ranges over $0$ to $p_i - 1$ we have $g^{\alpha p}$ ranges over all non zero residues modulo $p_i$. Now choose $\alpha$ such that $g^{\alpha p} \equiv -p \pmod{p_i}$. Now choose $m$ such that $m \equiv g^{\alpha} \pmod{p_i}$. It is clear that this $m$ works. So we are done. $\blacksquare$